vectors
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Vectors
Introduction
• In this chapter you will learn about Vectors
• You will have seen vectors at GCSE level, this chapter focuses on using them to solve problems involving SUVAT equations and forces
• Sometimes using vectors offers an easier alternative to regular methods
• Vectors are used in video games in the movement of characters and by engineers in the design of buildings, bridges and other structures
Teachings for Exercise 6A
VectorsYou can use vectors to describe
displacements
A vector has both direction and magnitude
For example:
An object is moving north at 20ms-1
A horizontal force of 7N
An object has moved 5m to the left
These are all vectors. A scalar quantity would be something such as:
A force of 10N
(It is scalar since it has no direction)
6A
Vectors have both direction and magnitude!
VectorsYou can use vectors to describe
displacements
A girl walks 2km due east from a fixed point O, to A, and then 3km due south
from A to a point B. Describe the displacement of B from O.
Start, as always, with a diagram!
To describe the displacement you need the distance from O as well as the
direction (as a bearing)
Remember bearings are always measured from north!
“Point B is 3.61km from O on a bearing of 146˚”
6A
2km
3km
O A
B
θ
N
Describing the displacement
The distance – use Pythagoras’ Theorem
𝑐=√𝑎2+𝑏2𝑐=√22+32𝑐=3.61𝑘𝑚The bearing – use Trigonometry to find angle θ
𝑇𝑎𝑛𝜃=𝑂𝑝𝑝𝐴𝑑𝑗
𝑇𝑎𝑛𝜃=32
𝜃=56.3 ˚𝐵𝑒𝑎𝑟𝑖𝑛𝑔=146 ˚
Sub in a and b
Calculate
Sub in opp and adj
Use inverse Tan
Bearings are measured from north. Add the north line and
add 90˚
Opp
Adj
56.3˚
VectorsYou can use vectors to describe
displacements
In an orienteering exercise, a cadet leaves the starting point S and walks 15km on a
bearing of 120˚ to reach A, the first checkpoint.
From A he then walks 9km on a bearing of 240˚ to the second checkpoint, and point
B.
From B he then returns directly to S. Describe the displacement of S from B.
Start with a diagram!
We need the distance B to S and the bearing from B to S as well. You will need to use angle rules you have
learnt pre A-level.
N
NS
A
B
15km
9km
120°
240°
You can use interior angles to find an angle
in the triangle Interior angles add up
to 180°The missing angle
next to 240 is 60°The angle inside the
triangle must also be 60°
60°
60°
Finding the distance B to S
𝑎2=𝑏2+𝑐2−2𝑏𝑐𝐶𝑜𝑠𝐴𝑎2=152+92−2 (15×9)𝐶𝑜𝑠60
𝑎2=171𝑎=13.1𝑘𝑚
Sub in values
a
b
c
Work out
Square root
13.1km
VectorsYou can use vectors to describe
displacements
In an orienteering exercise, a cadet leaves the starting point S and walks 15km on a
bearing of 120˚ to reach A, the first checkpoint.
From A he then walks 9km on a bearing of 240˚ to the second checkpoint, and point
B.
From B he then returns directly to S. Describe the displacement of S from B.
Start with a diagram!
We need the distance B to S and the bearing from B to S as well. You will need to use angle rules you have
learnt pre A-level.
N
NS
A
B
15km
9km
120°
240°60°
60°13.1km
N
θ
Finding the bearing from B to S Show the bearing at B It can be split into 2
sections, one of which is 180°
Find angle θ inside the triangle
𝑆𝑖𝑛𝜃9
=𝑆𝑖𝑛6013.1
𝑆𝑖𝑛θ=𝑆𝑖𝑛 6013.1
×9
𝜃=36.6 °
𝑆𝑖𝑛𝐴𝑎
=𝑆𝑖𝑛𝐵𝑏
Sub in values
Rearrange
Calculate θ
156.6° 180°
You can now use Alternate angles
to find the unknown part of
the bearing
Add on 180°
The bearing is 336.6°S is 13.1km from B on a bearing of 337°
A
a
bB
36.6°
Teachings for Exercise 6B
VectorsYou can add and represent
vectors using line segments
A vector can be represented as a directed line segment
Two vectors are equal if they have the same magnitude and
direction
Two vectors are parallel if they have the same direction
You can add vectors using the triangle law of addition
6B
A C
B
a 3a
𝐴𝐶=𝐴𝐵+𝐵𝐶
VectorsYou can add and represent
vectors using line segments
OACB is a parallelogram. The points P, Q, M and N are the
midpoints of the sides.
OA = a
OB = b
Express the following in terms of a and b.
a) OC b) AB c) QC
d) CN e) QN
6B
M
B
P
O
ND
QA C
b
a
a + b b - a 1/2b
-1/2a 1/2b - 1/2a
What can you deduce about AB and QN, looking at the vectors?
�⃗�𝐵=𝒃−𝒂 �⃗�𝑁=12𝒃−
12𝒂
�⃗�𝑁=12(𝒃−𝒂)
QN is a multiple of AB, so they are
parallel!
VectorsYou can add and represent
vectors using line segments
In triangle OAB, M is the midpoint of OA and N divides AB in the
ratio 1:2.
OM = a
OB = b
Express ON in terms of a and b
6B
A
BO
MN
a
b
a1
2
�⃗�𝑁=�⃗�𝐴+ �⃗�𝑁
�⃗�𝐴=2𝑎
�⃗�𝑁=13�⃗�𝐵
Use the ratio. If N divides AB in the ratio 1:2, show this on the diagram You can see now that AN is one-third of AB We therefore need to know AB To get from A to B, use AO + OB
�⃗�𝐵= �⃗�𝑂+�⃗�𝐵
�⃗�𝐵=−2𝒂+𝒃
�⃗�𝑁=−23𝒂+13𝒃
Sub in AO and OB
AN = 1/3AB
�⃗�𝑁=2𝒂+−23𝒂+13𝒃
�⃗�𝑁=43𝒂+13𝒃
Sub in values
Simplify
VectorsYou can add and represent
vectors using line segments
OABC is a parallelogram. P is the point where OB and AC intersect.
The vectors a and c represent OA and OC respectively.
Prove that the diagonals bisect each other.
If the diagonals bisect each other, then P must be the
midpoint of both AC and OB…
Try to find a way to represent OP in different ways…
(make sure you don’t ‘accidentally’ assume P is the
midpoint – this is what we need to prove!) 6B
P
O
A B
C
a
c
One way to get from O to P Start with OB
�⃗�𝐵=𝑎+𝑐
�⃗�𝑃= λ(𝑎+𝑐 )
OP is parallel to OB so is a multiple of (a + c)
We don’t know how much for now, so can use λ (lamda) to represent the unknown quantity
c
�⃗�𝑃= λ(𝑎+𝑐 )
VectorsYou can add and represent
vectors using line segments
OABC is a parallelogram. P is the point where OB and AC intersect.
The vectors a and c represent OA and OC respectively.
Prove that the diagonals bisect each other.
If the diagonals bisect each other, then P must be the
midpoint of both AC and OB…
Try to find a way to represent OP in different ways…
(make sure you don’t ‘accidentally’ assume P is the
midpoint – this is what we need to prove!) 6B
P
O
A B
C
a
c
Another way to get from O to P Go from O to A, then A to P We will need AC first…
c
�⃗�𝑃= λ(𝑎+𝑐 )
-a
�⃗�𝐶=𝑐−𝑎
�⃗�𝑃=𝜇(𝑐−𝑎)
�⃗�𝑃=�⃗�𝐴+ �⃗�𝑃
�⃗�𝑃=𝑎+𝜇(𝑐−𝑎)
AP is parallel to AC so is a multiple of it. Use a different symbol (usually μ, ‘mew’, for this multiple)Now we have another way to get from O to P
Sub in vectors
�⃗�𝑃=𝑎+𝜇(𝑐−𝑎)
VectorsYou can add and represent
vectors using line segments
OABC is a parallelogram. P is the point where OB and AC intersect.
The vectors a and c represent OA and OC respectively.
Prove that the diagonals bisect each other.
If the diagonals bisect each other, then P must be the
midpoint of both AC and OB…
Try to find a way to represent OP in different ways…
(make sure you don’t ‘accidentally’ assume P is the
midpoint – this is what we need to prove!) 6B
P
O
A B
C
a
�⃗�𝑃= λ(𝑎+𝑐 )�⃗�𝑃=𝑎+𝜇(𝑐−𝑎)
�⃗�𝑃= λ(𝑎+𝑐 ) �⃗�𝑃=𝑎+𝜇(𝑐−𝑎)
λ (𝑎+𝑐 )=𝑎+𝜇(𝑐−𝑎)
λ 𝑎+ λ𝑐=𝑎+𝜇𝑐−𝜇𝑎λ 𝑎+ λ𝑐=(1−𝜇)𝑎+𝜇𝑐
As these represent the same vector, the expressions must be equal!
Multiply out brackets
Factorise the ‘a’ terms on the right side
Now compare sides – there must be the same number of ‘a’s and ‘c’s on each
λ=1−𝜇 λ=𝜇λ=1−λλ=0.5μ=0.5
Sub 2nd equation into the first
They are equal
Rearrange and solve So P is halfway along OB and AC and hence the lines bisect each
other!
Teachings for Exercise 6C
VectorsYou can describe vectors using
the i, j notation
A unit vector is a vector of length 1. Unit vectors along Cartesian (x, y) axes are usually denoted by i and j
respectively.
You can write any two-dimensional vector in the form ai + bj
Draw a diagram to represent the vector -3i + j
6C
O
(0,1)
(1,0)
j
i
A B
C
5i
2j5i + 2j �⃗�𝐶= �⃗�𝐵+ �⃗�𝐶
�⃗�𝐶=5 𝒊+2 𝒋
-3i
j-3i + j
Teachings for Exercise 6D
VectorsYou can solve problems with vectors written using the i, j
notation
When vectors are written in terms of the unit vectors i and j you can add them together by adding the
terms in i and j separately. Subtraction works in a similar way.
Given that:p = 2i + 3jq = 5i + j
Find p + q in terms of i and j
6D
𝒑+𝒒=¿(2 𝒊+3 𝒋)+(5 𝒊+ 𝒋)
𝒑+𝒒=¿7 𝒊+4 𝒋
Add the i terms and j terms separately
VectorsYou can solve problems with vectors written using the i, j
notation
When vectors are written in terms of the unit vectors i and j you can add them together by adding the
terms in i and j separately. Subtraction works in a similar way.
Given that:a = 5i + 2jb = 3i - 4j
Find 2a – b in terms of i and j
6D
2 2(5 𝒊+2 𝒋 )−(3 𝒊−4 𝒋) Multiply out the bracket
2 10 𝒊+4 𝒋−(3 𝒊−4 𝒋)
2 10 𝒊+4 𝒋−3 𝒊+4 𝒋2 7 𝒊+8 𝒋
Careful with the subtraction here!
Group terms…
VectorsYou can solve problems with vectors written using the i, j
notation
When a vector is given in terms of the unit vectors i and j, you can
find its magnitude using Pythagoras’ Theorem.
The magnitude of vector a is written as |a|
Find the magnitude of the vector: 3i – 7j
6D
3i
-7j
3i - 7j
|𝒗|=√32+(−7 )2
|𝒗|=√58 (3sf)
Put in the values from the vectors and
calculateRound if necessary
VectorsYou can solve problems with vectors written using the i, j
notation
You can also use trigonometry to find an angle between a vector
and the axes
Find the angle between the vector -4i + 5j and the positive x-axis
Draw a diagram
6D
-4ix
θ5j
y
Opp
Adj
𝑇𝑎𝑛𝜃=𝑂𝑝𝑝𝐴𝑑𝑗
𝑇𝑎𝑛𝜃=54
𝜃=51.3 °
𝐴𝑐𝑡𝑢𝑎𝑙𝑎𝑛𝑔𝑙𝑒=128.7 °
Sub in values
Inverse Tan
The angle we want is between the vector and the positive x-axis
Subtract θ from 180°
51.3°
VectorsYou can solve problems with vectors written using the i, j
notation
Given that:
a = 3i - jb = i + j
Find µ if a + µb is parallel to 3i + j
Start by calculating a + µb in terms of a, b and µ
6D
𝒂+𝜇𝒃=¿(3 𝒊− 𝒋)+𝜇(𝒊+ 𝒋)
¿3 𝒊− 𝒋+𝜇𝒊+𝜇 𝒋
¿3 𝒊+𝜇𝒊− 𝒋+𝜇 𝒋
¿ (3+𝜇) 𝒊+(−1+𝜇) 𝒋
As the vector must be parallel to 3i + j, the i term must be 3 times the j term!
3+𝜇=¿3 (−1+𝜇)3+𝜇=¿−3+3𝜇6=¿2𝜇3=¿𝜇
Multiply out the brackets
Divide by 2
𝜇=3
Move the i and j terms together
Factorise the terms in i and j
Multiply out the bracket
Subtract µ, and add 3
¿2(3 𝒊+ 𝒋)
VectorsYou can solve problems with vectors written using the i, j
notation
Given that:
a = 3i - jb = i + j
Find µ if a + µb is parallel to 3i + j
Start by calculating a + µb in terms of a, b and µ
6D
𝜇=3
To show that this works…
𝒂+𝜇𝒃𝒂+3𝒃¿ (3 𝒊− 𝒋)+3 (𝒊+ 𝒋)
¿3 𝒊− 𝒋+3 𝒊+3 𝒋¿6 𝒊+2 𝒋
Multiply out the brackets
We now know µ
Group terms
Factorise
You can see that using the value of µ = 3, we get a vector which is parallel
to 3i + j
Teachings for Exercise 6E
VectorsYou can express the velocity of
a particle as a vector
The velocity of a particle is a vector in the direction of motion. The magnitude of the vector is its
speed. Velocity is usually represented by v.
A particle is moving with constant velocity given by:
v = (3i + j) ms-1
Find:a) The speed of the particle
b) The distance moved every 4 seconds
6E
Finding the speed
The speed of the particle is the magnitude of the vector Use Pythagoras’ Theorem
3i
j3i + j
|𝑣|=√32+12|𝑣|=3.16𝑚𝑠− 1
Finding the distance travelled every 4 seconds
Use GCSE relationships Distance = Speed x Time
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒=𝑠𝑝𝑒𝑒𝑑×𝑡𝑖𝑚𝑒𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒=3.16×4𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒=12.6𝑚
Sub in values (use the exact speed!)
Calculate
Calculate
Teachings for Exercise 6F
VectorsYou can solve problems
involving velocity, time, mass and forces (as in earlier
chapters) by using vector notation
If a particle starts from the point with position vector r0 and moves with constant velocity v, then its
displacement from its initial position at time t is given by:
6F
𝒓=𝒓 0+𝒗 𝑡
Final positio
n
Starting position
Velocity
Time
A particle starts from the point with position vector (3i + 7j) m and then moves constant
velocity (2i – j) ms-1. Find the position vector of the particle 4 seconds later.
(a position vector tells you where a particle is in relation to the origin O)
𝒓 0=(3 𝒊+7 𝒋)𝒗=(2 𝒊− 𝒋)𝑡=4
𝒓=𝒓 0+𝒗 𝑡
𝒓=(3 𝒊+7 𝒋 )+(2 𝒊− 𝒋 )(4 )
𝒓=3 𝒊+7 𝒋+8 𝒊−4 𝒋𝒓=11𝒊+3 𝒋
Sub in values
Multiply/remove brackets
Simplify
VectorsYou can solve problems
involving velocity, time, mass and forces (as in earlier
chapters) by using vector notation
If a particle starts from the point with position vector r0 and moves with constant velocity v, then its
displacement from its initial position at time t is given by:
6F
𝒓=𝒓 0+𝒗 𝑡
Final positio
n
Starting position
Velocity
Time
A particle moving at a constant velocity, ‘v’, and is at the point with position vector (2i + 4j) m at time t
= 0. Five seconds later the particle is at the point with position vector (12i + 16j) m. Find the velocity
of the particle.𝒓 0=(2 𝒊−4 𝒋 )𝒓=(12 𝒊+16 𝒋 )𝑡=5
𝒓=𝒓 0+𝒗 𝑡
(12 𝒊+16 𝒋)=(2 𝒊−4 𝒋 )+(𝒗 )(5)Sub in values
12 𝒊+16 𝒋=2 𝒊−4 𝐣+5 𝒗10 𝒊+20 𝒋=5 𝒗2 𝒊+4 𝒋=𝒗
The velocity of the particle is (2i + 4j) ms-1
Deal with the brackets!
Subtract 2i and add 4j
Divide by 5
VectorsYou can solve problems involving
velocity, time, mass and forces (as in earlier chapters) by using vector
notation
At time t = 0, a particle has position vector 4i + 7j and is moving at a speed of
15ms-1 in the direction 3i – 4j. Find its position vector after 2 seconds.
You have to be careful here, you have been given the speed of the particle.
The direction vector does not necessarily have the same speed
Find the speed of the direction vector as it is given in the question
Then ‘multiply up’ to get the required speed (we need 15ms-1, not 5ms-1)
Multiplying the vectors will allow you to use the correct velocity
6F
𝒓=𝒓 0+𝒗 𝑡
3i
-4j3i – 4j
𝑆𝑝𝑒𝑒𝑑=√32+(−4)2
𝑆𝑝𝑒𝑒𝑑=5𝑚𝑠−1
9i
-12j9i – 12j
5ms-1 15ms-1
Multiply all vectors by 3
𝑆𝑝𝑒𝑒𝑑=15𝑚𝑠−1
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦=(9 𝒊−12 𝒋 )𝑚𝑠−1
Calculate
We can use the vectors as the
velocity
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦=(9 𝒊−12 𝒋 )𝑚𝑠−1
VectorsYou can solve problems involving
velocity, time, mass and forces (as in earlier chapters) by using vector
notation
At time t = 0, a particle has position vector 4i + 7j and is moving at a speed of
15ms-1 in the direction 3i – 4j. Find its position vector after 2 seconds.
You have to be careful here, you have been given the speed of the particle.
The direction vector does not necessarily have the same speed
Find the speed of the direction vector as it is given in the question
Then ‘multiply up’ to get the required speed (we need 15ms-1, not 5ms-1)
Multiplying the vectors will allow you to use the correct velocity
6F
𝒓=𝒓 0+𝒗 𝑡
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦=(9 𝒊−12 𝒋 )𝑚𝑠−1
𝒓 0=(4 𝒊+7 𝒋)𝒗=(9 𝒊−12 𝒋) 𝑡=2
𝒓=𝒓 0+𝒗 𝑡
𝒓=(4 𝒊+7 𝒋)+(9 𝒊−12 𝒋 )(2)
𝒓=4 𝒊+7 𝒋+18 𝒊−24 𝒋𝒓=22 𝒊−17 𝒋
Sub in values
‘Deal with’ the brackets
Group terms
VectorsYou can solve problems
involving velocity, time, mass and forces (as in earlier
chapters) by using vector notation
You can also solve problems involving acceleration by using:
v = u + at
Where v, u and a are all given in vector form.
Particle P has velocity (-3i + j) ms-1 at time t = 0. The particle moves
along with constant acceleration a = (2i + 3j) ms-2. Find the speed of
the particle after 3 seconds.
6F
𝑠=?𝒖=(−3 𝒊+ 𝒋)𝒗=? 𝒂=(2 𝒊+3 𝒋 )𝑡=3
𝒗=𝒖+𝒂𝑡
𝒗=(−3 𝒊+ 𝒋)+(2 𝒊+3 𝒋)(3)
𝒗=−3 𝒊+ 𝒋+6 𝒊+9 𝒋
𝒗=3 𝒊+10 𝒋
𝑆𝑝𝑒𝑒𝑑=√32+102𝑆𝑝𝑒𝑒𝑑=10.4𝑚𝑠−1
Sub in values
‘Deal with’ the brackets
Group terms
Remember this is the velocity, not the speed!
Calculate!
VectorsYou can solve problems
involving velocity, time, mass and forces (as in earlier
chapters) by using vector notation
A force applied to a particle has both a magnitude and a direction, so force is a vector. The force will cause the particle
to accelerate.
Remember from chapter 3:
F = ma
A constant force, FN, acts on a particle of mass 2kg for 10 seconds. The particle is initially at rest, and 10
seconds later it has a velocity of (10i – 24j) ms-1. Find F.
We need to find a first…
6F
𝑠=?𝒖=0𝒗=(10 𝒊−24 𝒋 )𝒂=?𝑡=10
𝒗=𝒖+𝒂𝑡(10 𝒊−24 𝒋 )=0+𝒂(10)
10 𝒊−24 𝒋=10𝒂𝒊−2.4 𝒋=𝒂
Sub in values
‘Tidy up’
Divide by 10
𝑭=𝑚𝒂𝑭=(2)(𝒊−2.4 𝒋)
𝑭=(2 𝒊−4.8 𝒋 ) 𝑁
Sub in values
Calculate
Teachings for Exercise 6G
VectorsYou can use vectors to solve
problems about forces
If a particle is resting in equilibrium, then the resultant of all the forces acting on it is zero.
The forces (2i + 3j), (4i – j), (-3i + 2j) and (ai + bj) are acting on a particle which is in equilibrium.
Calculate the values of a and b.
Set the sum of all the vectors equal to 0
6G
(2 𝒊+3 𝒋 )+(4 𝒊− 𝒋 )+ (−3 𝒊+2 𝒋 )+ (𝑎𝒊+𝑏 𝒋 )=0
(3 𝒊+𝟒 𝒋 )+(𝑎 𝒊+𝑏 𝒋 )=0
3+𝑎=0𝑎=−3
4+𝑏=0𝑏=−4
Group together the
numerical terms
The ‘i’ terms must sum to
0
The ‘j’ terms must sum to
0
VectorsYou can use vectors to solve
problems about forces
If several forces are involved in a question a good starting point is to
find the resultant force.
The following forces:
F1 = (2i + 4j) N
F2 = (-5i + 4j) N
F3 = (6i – 5j) N
all act on a particle of mass 3kg. Find the acceleration of the
particle.
Start by finding the overall resultant force.
6G
𝑅𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝐹𝑜𝑟𝑐𝑒=𝑭 1+𝑭 2+𝑭 3
¿ (2 𝒊+4 𝒋 )+(−5 𝒊+4 𝒋 )+(6 𝒊−5 𝒋)
¿ (3 𝒊+3 𝒋 )
𝑭=𝑚𝒂
(3 𝒊+3 𝒋)=3𝒂
𝒊+ 𝒋=𝒂
The acceleration is (i + j) ms-2
Sub in values
Group up
Sub in the resultant force, and the mass
Divide by 3
VectorsYou can use vectors to solve
problems about forces
A particle P, of weight 7N is suspended in equilibrium by two light inextensible strings attached to P and
to points A and B as shown in the diagram.
Line AB is horizontal. Find the tensions in the two strings
Draw a sketch of the forces acting on P
These can be rearranged into a triangle of forces
(the reason being, if the particle is in equilibrium then the overall force is
zero – ie) The particle ends up where it started)
You will now need to work out the angles in the triangle… 6G
A B
P
30° 40°
P
TA TB
7NTA
TB
7N
These are the forces acting on P
These are the forces rearranged as a
triangle
7N
VectorsYou can use vectors to solve
problems about forces
A particle P, of weight 7N is suspended in equilibrium by two
light inextensible strings attached to P and to points A and B as
shown in the diagram.
Line AB is horizontal. Find the tensions in the two strings
You will now need to work out the angles in the triangle…
Consider the original diagram, you could work out more
angles on it as shown, some of which correspond to our
triangle of forces…
6G
TA
TB
7NThe angle between 7N
and TA is 60°
A B
P
30° 40°
7N
50°60°
60°
50°
70°
The angle between 7N and TB is 50°
(It is vertically opposite on our triangle of
forces)The final angle can be worked out from the
triangle of forces alone
Now we can calculate the tensions!
VectorsYou can use vectors to solve
problems about forces
A particle P, of weight 7N is suspended in equilibrium by two
light inextensible strings attached to P and to points A and B as
shown in the diagram.
Line AB is horizontal. Find the tensions in the two strings
To calculate the tensions you can now use the Sine rule
(depending on the information given, you may have to use the
Cosine rule instead!)
6G
TA
TB
7N
A B
P
30° 40°
7N
50°60°
60°
50°
70°
𝑇 𝐴
𝑆𝑖𝑛50= 7𝑆𝑖𝑛70
𝑇 𝐴=7
𝑆𝑖𝑛70×𝑆𝑖𝑛50
𝑇 𝐴=5.71𝑁
Multiply by Sin50
Calculate
𝑇 𝐴=5.71𝑁
VectorsYou can use vectors to solve
problems about forces
A particle P, of weight 7N is suspended in equilibrium by two
light inextensible strings attached to P and to points A and B as
shown in the diagram.
Line AB is horizontal. Find the tensions in the two strings
To calculate the tensions you can now use the Sine rule
(depending on the information given, you may have to use the
Cosine rule instead!)
6G
TA
TB
7N
A B
P
30° 40°
7N
50°60°
60°
50°
70°
𝑇 𝐵
𝑆𝑖𝑛60= 7𝑆𝑖𝑛70
𝑇 𝐵=7
𝑆𝑖𝑛70×𝑆𝑖𝑛 60
𝑇 𝐵=6.45𝑁
Multiply by Sin60
Calculate
𝑇 𝐴=5.71𝑁
𝑇 𝐵=6.45𝑁
Teachings for Exercise 6H(the mixed exercise – essenti al!)
VectorsYou need to be able to solve worded
problems in practical contexts
The mixed exercise in this chapter is very important as it contains questions in
context, the type of which are often on exam papers
6H
VectorsYou need to be able to solve worded
problems in practical contexts
A ship S is moving with constant velocity (–2.5i + 6j) kmh–1. At time 1200, the position vector of S relative to a fixed
origin O is (16i + 5j) km.
Find:(a) the speed of S
(b) the bearing on which S is moving.
The ship is heading directly towards a submerged rock R. A radar tracking
station calculates that, if S continues on the same course with the same speed, it
will hit R at the time 1500.
(c) Find the position vector of R.
6H
-2.5i
6j
The speed of S
𝑆𝑝𝑒𝑒𝑑=√(−2.5)2+62
𝑆𝑝𝑒𝑒𝑑=6.5𝑘𝑚h−1
Use Pythagoras’
Theorem
Calculate
6.5 kmh-1
N
θ180°
The bearing on which S is travelling
𝑇𝑎𝑛𝜃=62.5
Find angle θ
𝜃=67.4 °
𝐵𝑒𝑎𝑟𝑖𝑛𝑔=337 °
Use Tan = Opp/Adj
Calculate
Consider the north line and
read clockwise…
337°
67.4°
VectorsYou need to be able to solve worded
problems in practical contexts
A ship S is moving with constant velocity (–2.5i + 6j) kmh–1. At time 1200, the position vector of S relative to a fixed
origin O is (16i + 5j) km.
Find:(a) the speed of S
(b) the bearing on which S is moving.
The ship is heading directly towards a submerged rock R. A radar tracking
station calculates that, if S continues on the same course with the same speed, it
will hit R at the time 1500.
(c) Find the position vector of R.
6H
6.5 kmh-1
337°
𝒓 0=(16 𝒊+5 𝒋)
𝒓=𝒓 0+𝒗 𝑡
𝒗=(−2.5 𝒊+6 𝒋)𝒕=3
𝒓=(16 𝒊+5 𝒋)+(−2.5 𝒊+6 𝒋)(3)
𝒓=16 𝒊+5 𝒋−7.5 𝒊+18 𝒋𝒓=8.5 𝒊+23 𝒋
Sub in values
‘Deal with’ the brackets
Group terms
𝑹=8.5 𝒊+23 𝒋
VectorsYou need to be able to solve worded
problems in practical contexts
A ship S is moving with constant velocity (–2.5i + 6j) kmh–1. At time 1200, the position vector of S relative to a fixed
origin O is (16i + 5j) km.
The tracking station warns the ship’s captain of the situation. The captain
maintains S on its course with the same speed until the time is 1400. He then changes course so that S moves due north at a constant speed of 5 km h–1.
Assuming that S continues to move with this new constant velocity.
Find:
(d) an expression for the position vector of the ship t hours after 1400,
(e) the time when S will be due east of R,(f) the distance of S from R at the time
1600 6H
Find the position vector of the ship at 1400
𝒓 0=(16 𝒊+5 𝒋)
𝒓=𝒓 0+𝒗 𝑡
𝒗=(−2.5 𝒊+6 𝒋)𝒕=2
𝒓=(16 𝒊+5 𝒋)+(−2.5 𝒊+6 𝒋)(2)
𝒓=16 𝒊+5 𝒋−5 𝒊+12 𝒋𝒓=11𝒊+17 𝒋
Sub in values
‘Deal with’ the brackets
Group terms
So at 1400 hours, the ship is at position vector (11i + 17j)
𝑹=8.5 𝒊+23 𝒋
VectorsYou need to be able to solve worded
problems in practical contexts
A ship S is moving with constant velocity (–2.5i + 6j) kmh–1. At time 1200, the position vector of S relative to a fixed
origin O is (16i + 5j) km.
The tracking station warns the ship’s captain of the situation. The captain
maintains S on its course with the same speed until the time is 1400. He then changes course so that S moves due north at a constant speed of 5 km h–1.
Assuming that S continues to move with this new constant velocity.
Find:
(d) an expression for the position vector of the ship t hours after 1400,
(e) the time when S will be due east of R,(f) the distance of S from R at the time
1600 6H
At 1400 the ship is at (11i + 17j)
Find an expression for its position t hours after 1400 Use the same formula, with the updated information
𝒓 0=(11𝒊+17 𝒋) 𝒗=5 𝒋 𝑡=𝑡
𝒓=𝒓 0+𝒗 𝑡
𝒓=(11 𝒊+17 𝒋)+(5 𝒋)(𝑡)
𝒓=11𝒊+17 𝒋+5 𝑡 𝒋𝒓=11𝒊+(17+5 𝑡) 𝒋
𝑹=8.5 𝒊+23 𝒋Sub in values
‘Deal with’ the brackets
Factorise the j terms
𝒓=11𝒊+(17+5 𝑡) 𝒋
VectorsYou need to be able to solve worded
problems in practical contexts
A ship S is moving with constant velocity (–2.5i + 6j) kmh–1. At time 1200, the position vector of S relative to a fixed
origin O is (16i + 5j) km.
The tracking station warns the ship’s captain of the situation. The captain
maintains S on its course with the same speed until the time is 1400. He then changes course so that S moves due north at a constant speed of 5 km h–1.
Assuming that S continues to move with this new constant velocity.
Find:
(d) an expression for the position vector of the ship t hours after 1400,
(e) the time when S will be due east of R,(f) the distance of S from R at the time
1600 6H
𝑹=8.5 𝒊+23 𝒋
𝒓=11𝒊+(17+5 𝑡) 𝒋
Find the time when S will be due east of R
R S
8.5 𝒊+23 𝒋 11𝒊+(17+5 𝑡) 𝒋If S is due east of R, then their j terms must be equal!
23=17+5 𝑡6=5 𝑡1.2=𝑡
Subtract 17
Divide by 5
1.2 hours = 1 hour 12 minutes
So S will be due east of R at 1512 hours!
1512
VectorsYou need to be able to solve worded
problems in practical contexts
A ship S is moving with constant velocity (–2.5i + 6j) kmh–1. At time 1200, the position vector of S relative to a fixed
origin O is (16i + 5j) km.
The tracking station warns the ship’s captain of the situation. The captain
maintains S on its course with the same speed until the time is 1400. He then changes course so that S moves due north at a constant speed of 5 km h–1.
Assuming that S continues to move with this new constant velocity.
Find:
(d) an expression for the position vector of the ship t hours after 1400,
(e) the time when S will be due east of R,(f) the distance of S from R at the time
1600 6H
𝑹=8.5 𝒊+23 𝒋
𝒓=11𝒊+(17+5 𝑡) 𝒋
1512
Find the distance of S from R at the time 1600 Find where S is at 1600 hours…
𝒓=11𝒊+(17+5 𝑡) 𝒋
𝒓=11𝒊+(17+5(2)) 𝒋
𝒓=11𝒊+27 𝒋
Sub in t = 2 (1400 – 1600
hours)
Simplify/calculate
So the position vectors of the rock and the ship at 1600 hours are:
𝑹=8.5 𝒊+23 𝒋 𝑺=11𝒊+27 𝒋To calculate the vector between them, calculate S - R
(11𝒊+27 𝒋 )−(8.5 𝒊+23 𝒋)¿2.5 𝒊+4 𝒋
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒=√2.52+42
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒=4.72𝑘𝑚Calculate
Now use Pythagoras’ Theorem to work out the distance
4.72𝑘𝑚
Summary
• We have seen how to use vectors in problems involving forces and SUVAT equations
• We have also seen how to answer multi-part worded questions
• It is essential you practice the mixed exercise in this chapter
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